It's possible to solve this problem without using trigonometry.The key is auxiliary construction:

Geometry problem solving is one of the most challenging skills for students to learn. When a problem requires auxiliary construction, the difficulty of the problem increases drastically, perhaps because deciding which construction to make is an ill-structured problem. By “construction,” we mean adding geometric figures (points, lines, planes) to a problem figure that wasn’t mentioned as "given."

http://i1237.photobucket.com/albums/ff480/Evan_Liang/DoubleangleProblem2.jpgA much easier way using angle bisector and similarityconstruct a reflection of △ABE △ABKalong the line of reflection AB∴△ABK≅△ABE∴∠KAB=∠BAE=a,AK=AE,KB=BE∵∠EAC=2a=∠KAE∴AE is the angle bisector of ∠KAC∴KE/AK=CE/AC=2BE/AEAE/AC=2BE/CE ..................(1)∵∠AED=∠ECA,∠EAC=∠EAC (reflexive prop.)∴△ADE∼△AEC (AA similarity)∴DE/CE=AE/AC ...................(2)Use transitive Prop. between (1) and (2)∴DE/CE=2BE/CEBE=2BE

Start like the other solutions and reflect ABE to a new triangle ABF. Then add point G on the line AC such that AGE = 90 - atri(EDG) is then isosceles and EG = ED. In addition tri(AEG) is isosclese with AE = AG. So tri(AEG) is congruent to tri(AFE) via SAS and FE = EG = ED = 2BE from that congruence.